Test: Rank of a Matrix & LU Decomposition - Civil Engineering (CE) MCQ

Given:

|A| = 1 x [6 x 2 - 2 x 7] - 3 × [4 x 2 - 1 × 7] + 5 x [4 x 2 - 1 x 6]
⇒ |A| = -2 - 3 + 10 = 5
So, |A| ≠ 0
The rank of A = 3

R₁ → -R₁

R₂ → R₂ - 2R₁; R₃ → R₃ - 3R₁; R₄ → R₄ - 5R₁

R₄ → R₄ - 2R₂; R₃ → R₃ - 2R₂

So, r[A] = 2

We must first complete the first step of forward elimination

First step: Multiply Row 1 by 10/25 = 0.4 and subtract the results from Row 2

Multiply Row 1 by 8/25 = 0.32,  and subtract the results from Row 3

To find ℓ₂₁ and ℓ₃₁ , what multiplier was used to make the a₂₁ and a₃₁ elements zero in the first step of forward elimination using the Naïve Gauss elimination method? They are
ℓ₂₁ = 0.4
ℓ₃₁ = 0.32
To find ℓ₃₂, what multiplier would be used to make the a₃₂ element zero? Remember the a₃₂ element is made zero in the second step of forward elimination.
So

Hence

Given:

Calculation of rank of the given matrix A:
Interchanging R₁ and R₂

The given matrix has rank 2.

∵ the matrix has a dimension of 3, and the rank is 2, therefore the rank of adjoint is 1.

Given

Number of non-zero rows = rank of the matrix = 2

∴ The rank of the following matrix is 2.

Given

Apply now transformation, R₄ → R₄ + R₁, we get,

R₂ ↔ R₃

This matrix is in triangular form with last row = 0,
So, Rank = 4

First step: The first step of forward elimination does not need to be conducted as 21 a and a31 are already zero.
Second step: Multiply Row 2 by 12/8 = 1.5 and subtract the results from Row 3

Thus

|X| = 0 (72 – 80) – (–1) (64 – 80) – 2(64 – 72) 
∴ |X| = 0 + (–16) + 16 = 0
Determinant of 3 × 3 matrix is zero.
∴ Rank(X) ≠ 3

Determinant of 2 × 2 matrix is nonzero. So, rank of X is 2.

Concept:

Properties of Rank: Rank of a matrix is the number of independent rows in the given matrix. Given two square matrices A and B of order n × n, we have following properties:

• 1. Rank of product of A and B i.e. Rank (AB) ≥ Rank (A) + Rank (B) – order of square matrix
• 2. Rank of sum of A and B i.e. Rank (A + B) ≤ Rank (A) + Rank (B).

Properties of Determinant: Given two square matrices A and B of order n × n and their determinants Det(A) and Det(B) respectively, determinant of their product i.e Det( AB) = Det(A) * Det(B). However, the same does not hold for the addition of the given matrices.

Example:

Consider two square matrices A and B each of order 2×2.

Rank of B = 2
Det (B) = 1

Rank (AB) = 2
Det (AB) = 5

Rank of A+B = 2

Det (A+B) = 12

Statement I is FALSE.
The rank of product matrix AB is 2. Product of Rank(A) and Rank(B) is: 2*2 = 4. Therefore, the rank of the product matrix is not equal to the product of the rank of individual matrices.

Statement II is TRUE. 
Det(AB)= 5 = Det(A) * Det(B)

Statement III is TRUE.
Rank(A + B) = 2. Sum of rank of A and B is: 2 + 2 = 4. Therefore, the relation: Rank (A + B) ≤ Rank (A) + Rank (B) holds true
Therefore, the rank of the addition matrix is less than or equal to the sum of the rank of the individual matrices.

Statement IV is FALSE.  
Det(A+B)= 12, which is greater than the sum of the determinants of individual matrices.

Given:

Replace R₂ with R₄ matrix becomes

Applying row transformations R₃ → R₃ + R₁ Now the matrix becomes.

Applying Row transformation for row 4: R₄ → R₄ - 2 R₂

Replace R₃ with R₄, the matrix becomes

Therefore Number of non zero rows is 3 
∴ The rank of the matrix is 3